DOI: 10.1515/auom-2016-0018
An. Şt. Univ. Ovidius Constanţa
Vol. 24(1),2016, 309–333
Matkowski theorems in the context of
quasi-metric spaces and consequences on
G-metric spaces
Erdal Karapınar, Antonio-Francisco Roldán-López-de-Hierro and
Bessem Samet
Abstract
In this paper, we prove the characterization of a Matkowski’s theorem in the setting of quasi-metric spaces. As a result, we observe that
some recent fixed point results in the context of G-metric spaces are
consequences of our main result.
1
Introduction
After the appearance of the Banach Contractive Mapping Principle in his
thesis in 1922, Fixed Point Theory has become one of the most useful tools in
Nonlinear Analysis due, mainly, to its applications. Many results have been
introduced in this field throughout the last ninety years.
In [1], Matkowski presented the following result.
Theorem 1.1 (Matkowski [1], Theorem 1). Let (X, d) be a complete metric
5
space, T : X → X, α : [0, ∞) → [0, ∞), and let γ(t) = α(t, t, t, 2t, 2t) for all
t ≥ 0. Suppose that
(1) α is nondecreasing with respect to each variable,
(2) lim (t − γ(t)) = ∞,
t→∞
Key Words: Fixed point, Contractivity condition, Quasi-metric space, G-metric space.
2010 Mathematics Subject Classification: Primary 47H10, 54H25; Secondary 47H05,
65J15.
Received:29.10.2014
Accepted:07.01.2015
309
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consequences on G-metric spaces
310
(3) lim γ k (t) = 0 for all t > 0,
k→∞
(4) for every x ∈ X, there is a positive integer n = n(x) such that, for all
y ∈ X,
d(T n(x) x, T n(x) y) ≤ α(d(x, T n x), d (x, T n y) , d (x, y) , d (T n x, y) , d (T n y, y)).
(1)
k
Then T has a unique fixed point a ∈ X and for each x ∈ X, lim T x = a.
k→∞
Only considering the function γ : [0, ∞) → [0, ∞), Matkowski obtained the
following consequence.
Corollary 1.1 (Matkowski [1], Theorem 2). Let (X, d) be a complete metric
space and let T : X → X be a mapping. Assume that there exists a nondecreasing function γ : [0, ∞) → [0, ∞) such that, for all t > 0, we have that
lim (t − γ(t)) = ∞ and lim γ k (t) = 0, and verifying that for each x ∈ X,
t→∞
k→∞
there is a positive integer n = n(x) such that
d(T n(x) x, T n(x) y) ≤ γ(d(x, y))
for all y ∈ X.
(2)
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, lim T k x =
k→∞
a.
Very recently, Gajić and Stojaković [2] proved the analog of Matkowski’s
theorem in the context of G-metric spaces as follows. We recall the classification of auxiliary functions given in [2]. Let ϕ : [0, ∞) → [0, ∞) be a
nondecreasing function. The additional properties that can be imposed on ϕ
are listed below:
(ϕ1 ) ϕ(0) = 0;
(ϕ2 ) ϕ(t) < t, for all t > 0;
(ϕ3 ) lim ϕk (t) = 0, for all t > 0;
k→∞
(ϕ4 ) if {ti } ⊂ [0, ∞) is a sequence such that ti+1 ≤ ϕ(ti ), then lim ti = 0;
i→∞
(ϕ5 ) for any y ≥ 0 there exists a t(y) ≥ 0, t(y) = sup{t ≤ y + ϕ(t)};
t≥0
(ϕ6 ) lim (t − ϕ(t)) = ∞;
t→∞
(ϕ7 )
∞
X
i=1
ϕi (t) < ∞, for all t > 0.
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311
Theorem 1.2 (Gajić and Stojaković [2], Theorem 2.1). Let (X, G) be a complete G-metric space, T : X → X, and ϕ : [0, ∞) → [0, ∞). If ϕ is nondecreasing mapping that satisfies either (ϕ3 )&(ϕ6 ) or (ϕ4 )&(ϕ5 ) and for each
x ∈ X there is a positive integer n = n(x) such that for all y ∈ X,
G(T n(x) x, T n(x) x, T n(x) y) ≤ ϕ(G(x, x, y)),
then T has a unique fixed point a ∈ X.
lim T k (x) = a.
Moreover, for each x ∈ X,
k→∞
In recent times, some authors have proved that many fixed point theorems
in the context of G-metric spaces can be deduced from existing results in the
context of quasi-metric spaces and/or metric spaces (see e.g. [3, 4, 5, 6, 7, 8]).
However, in [2], the authors claimed that their results cannot be observed from
the corresponding results in the frame of usual metric spaces or quasi metric
spaces via the techniques used in [3, 4].
This manuscript has two aims: on the one hand, we introduce two different conditions to prove the analog of Matkowski’s theorem in the setting
of quasi-metric spaces by verbatim; on the other hand, we particularize our
main results to the setting of G-metric spaces pointing out that Gajić and
Stojaković’s results can be easily derived from such consequences by using the
same techniques in [3, 4].
2
Preliminaries
Throughout this paper, N = {0, 1, 2, . . .} denotes the set of nonnegative integers and R denotes the set of all real numbers. First, let us recall the following
definitions, notations and basic results.
Definition 2.1. A quasi-metric on X is a function q : X × X → [0, ∞)
satisfying the following properties:
(q1 ) q(x, y) = 0 if and only if x = y;
(q2 ) q(x, y) ≤ q(x, z) + q(z, y) for any points x, y, z ∈ X.
In such a case, the pair (X, q) is called a quasi-metric space.
It is clear that any metric space is a quasi-metric space, but the converse
is not true. Now, we give the notions of convergence and completeness on
quasi-metric spaces.
Definition 2.2. Let (X, q) be a quasi-metric space, {xn } be a sequence in X,
and x ∈ X. We say that:
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• {xn } converges to x (and we denote it by {xn } → x) if
lim q(xn , x) = lim q(x, xn ) = 0;
n→∞
n→∞
• {xn } is a Cauchy sequence if for all ε > 0, there exists n0 ∈ N such that
q(xn , xm ) < ε for all n, m ≥ n0 .
The quasi-metric space is said to be complete if every Cauchy sequence is
convergent.
As q is not necessarily symmetric, some authors distinguished between
left/right Cauchy/convergent sequences and completeness.
Definition 2.3. (Jleli and Samet [4]) Let (X, q) be a quasi-metric space, {xn }
be a sequence in X, and x ∈ X. We say that:
• {xn } right-converges to x if limn→∞ q(xn , x) = 0;
• {xn } left-converges to x if limn→∞ q(x, xn ) = 0;
• {xn } is a right-Cauchy sequence if for all ε > 0 there exists n0 ∈ N such
that q(xn , xm ) < ε for all m > n ≥ n0 ;
• {xn } is a left-Cauchy sequence if for all ε > 0 there exists n0 ∈ N such
that q(xm , xn ) < ε for all m > n ≥ n0 .
Remark 2.1. A sequence {xn } in a quasi-metric space is Cauchy if, and only
if, it is left-Cauchy and right-Cauchy.
Definition 2.4. Let (X, q) be a quasi-metric space and let T : X → X be a
mapping. We say that T is
• right-continuous if {q(T xn , T u)} → 0 for all sequence {xn } ⊆ X and all
u ∈ X such that {q(xn , u)} → 0;
• left-continuous if {q(T u, T xn )} → 0 for all sequence {xn } ⊆ X and all
u ∈ X such that {q(u, xn )} → 0;
• T is continuous if {T xn } → T u for all sequence {xn } ⊆ X and all u ∈ X
such that {xn } → u.
Notice that if T is, at the same time, right-continuous and left-continuous,
then it is continuous.
With respect to conditions (ϕ1 ) to (ϕ7 ) introduced in the previous section,
Gajić and Stojaković demonstrated the following relationships.
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Lemma 2.1. ([2]) Let ϕ : [0, ∞) → [0, ∞) be a nondecreasing function. Then
(i) (ϕ3 ) ⇔ (ϕ4 ) ⇒ (ϕ2 ),
(ii) if ϕ is right continuous, then (ϕ2 ) ⇔ (ϕ3 ) ⇔ (ϕ4 ),
(iii) (ϕ7 ) ⇒ (ϕk ) ⇒ (ϕ2 ) ⇒ (ϕ1 ), where k ∈ {3, 4},
(iv) (ϕ5 ) ⇔ (ϕ6 ),
(v) (ϕ2 ) ; (ϕ3 ) and (ϕ2 ) ; (ϕ4 ),
(vi) (ϕ5 ) + (ϕ3 ) ; (ϕ7 ) and (ϕ6 ) + (ϕ3 ) ; (ϕ7 ),
(vii) (ϕ7 ) ; (ϕ5 ) and (ϕ7 ) ; (ϕ6 ).
Corollary 2.1. ([2]) A nondecreasing mapping ϕ : [0, ∞) → [0, ∞) verifies
(ϕ3 ) and (ϕ6 ) if, and only if, it satisfies (ϕ4 ) and (ϕ5 ).
3
The Matkowski’s theorem in the context of quasi-metric
spaces
In this section, we will prove two different analogs of the previous theorem in
the context of quasi-metric spaces. To be exact, we will need an additional
hypothesis on the contractive condition or in the quasi-metric space.
Let denote by FMat the family of all functions ϕ : [0, ∞) → [0, ∞) as in
Theorem 1.1, that is, ϕ is nondecreasing and it satisfies (ϕ3 ) and (ϕ6 ). By
Lemma 2.1, we have the following characterization.
Lemma 3.1. (Gajić and Stojaković [2]) A nondecreasing function ϕ : [0, ∞) →
[0, ∞) belongs to FMat if, and only if, it satisfies (ϕ4 ) and (ϕ5 ).
In fact, as (ϕ3 ) implies (ϕ2 ), we also have the following result.
Lemma 3.2. Suppose that ϕ ∈ FMat .
1. ϕ(t) < t
for all t > 0.
2. ϕ(t) ≤ t
for all t ≥ 0.
3. ϕ (0) = 0.
4. lim ϕk (t) = 0
k→∞
for all t ≥ 0.
5. ϕk is nondecreasing for all k ∈ N.
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M -symmetric quasi-metric spaces
In this subsection, we present a simple condition which will be able to consider
an extension of Matkowski’s theorem to quasi-metric spaces and to G-metric
spaces.
Definition 3.1. Given a positive real number M > 0, we will say that a
quasi-metric space (X, q) is M -symmetric if
q (y, x) ≤ M q (x, y)
for all x, y ∈ X.
If X is not reduced to a single point, we can find x, y ∈ X such that
q (x, y) > 0. Therefore, 0 < q (x, y) ≤ M q (y, x) ≤ M 2 q (x, y). Hence, M ≥ 1.
Example 3.1. Using M = 1, every metric space is a symmetric space. In
fact, a quasi-metric space is a metric space if, and only if, it is 1-symmetric.
Example 3.2. Let X = R and let define
(
2 (x − y) , if x ≥ y,
q(x, y) =
y − x,
if x < y.
Then (X, q) is a complete 2-symmetric quasi-metric space, but it is not a
metric space.
The main properties of M -symmetric quasi-metric spaces are listed in the
following result.
Lemma 3.3. Let (X, q) be an M -symmetric quasi-metric space, let {xn } ⊆ X
be a sequence and let x ∈ X. Then the following properties hold.
1. The following conditions are equivalent.
(a) {xn } right-converges to x.
(b) {xn } left-converges to x.
(c) {xn } converges to x.
2. If a sequence is right-convergent, then it is convergent, and its limit
coincide with its right-limit.
3. The following conditions are equivalent.
(a) {xn } is right-Cauchy.
(b) {xn } is left-Cauchy.
(c) {xn } is Cauchy.
4. If {yn } ⊆ X and {q (xn , yn )} → 0, then {q (yn , xn )} → 0.
Matkowski theorems in the context of quasi-metric spaces and
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315
The Matkowski’s theorem on M -symmetric quasi-metric spaces
In this subsection, we prove a version of Theorem 1.1 using the M -symmetry
of the quasi-metric space.
5
Definition 3.2. Given M ≥ 1 and two functions α : [0, ∞) → [0, ∞) and
γ : [0, ∞) → [0, ∞), we will say that (α, γ, M ) is a Matkowski’s triple if the
following conditions are fulfilled:
(P1 ) α is nondecreasing with respect to each variable and γ is nondecreasing;
(P2 ) α (t, t, t, (M + 1)t, (M + 1)t) ≤ γ(t) for all t ≥ 0;
(P3 ) lim (t − γ(t)) = ∞;
t→∞
(P4 ) lim γ k (t) = 0 for all t > 0.
k→∞
Notice that if (α, γ, M ) is a Matkowski’s triple, then γ ∈ FMat , and all
items of Lemma 3.2 are applicable.
Example 3.3. Suppose that, given M ≥ 1 and λ ∈ [0, 1), α and γ are defined
by
t4
t5
α (t1 , t2 , t3 , t4 , t5 ) = λ max t1 , t2 , t3 ,
,
,
M +1 M +1
γ(t) = α (t, t, t, (M + 1)t, (M + 1)t) = λ t.
Then (α, γ, M ) is a Matkowski’s triple.
The following one is the main result of the present manuscript.
Theorem 3.1. Let (X, q) be a complete M -symmetric quasi-metric space and
let T : X → X be a mapping. Suppose that there exists a Matkowski’s triple
(α, γ, M ) verifying the following property:
• for every x ∈ X, there is a positive integer n = n(x) such that, for all
y ∈ X,
q(T n(x) x, T n(x) y) ≤ α q (x, y) , q(x, T n(x) x),
q(x, T n(x) y), q(T n(x) x, y), q(T n(x) y, y) . (3)
Then T has a unique fixed point a ∈ X. Furthermore, for each x ∈ X,
lim T k x = a and T n(a) is continuous at a.
k→∞
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Remark 3.1. Notice that the mapping T is not assumed to be continuous.
The previous theorem improves Theorem 1.1 in three senses: (1) we only assume that (X, q) is a quasi-metric space; (2) we only suppose that
α (t, t, t, (M + 1)t, (M + 1)t) ≤ γ(t) for all t ≥ 0, but the equality is not necessary; (3) we also prove that T n(a) is continuous at a.
Our proof is based on the original Matkowski’s proof given in [1]. However,
some details are different. Then, for a better understanding, we divide the
proof in 14 steps.
Proof. Step 1. We claim that, for all x ∈ X, the set {q(x, T k x) : k ∈ N} is
bounded.
Let x ∈ X be arbitrary. By hypothesis, there exists a positive integer
n = n(x) such that (3) holds. Given an integer s ∈ {0, 1, 2, . . . , n(x) − 1}, we
are going to show that the set {q(x, T k n(x)+s x) : k ∈ N} is bounded (varying
s on {0, 1, 2, . . . , n(x) − 1}, we can conclude that Step 1 holds). We define
uk = q(x, T k n(x)+s x)
h = max{u0 , q(x, T
n(x)
for all k ∈ {0, 1, 2, . . .} ,
)}.
Due to (ϕ6 ), there exists c ∈ (0, ∞), with c > h, such that
t − ϕ(t) > h for all t ∈ [ c, ∞) .
(4)
Since c > h ≥ u0 , then
u0 < c.
Next, we claim that uk < c for all k ∈ N. On the contrary, assume that there
exists a positive integer j such that uj ≥ c but ui < c for all i < j. Notice that
uj−1 < c ≤ uj . Taking (3) into account together with the triangle inequality,
we derive that
uj = q(x, T j n(x)+s x) ≤ q(x, T n(x) x) + q(T n(x) x, T j n(x)+s x)
≤ h + q(T n(x) x, T n(x) T (j−1)n(x)+s x).
(5)
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Using the contractivity condition (3) with y = T (j−1)n(x)+s x,
q(T n(x) x, T n(x) T (j−1)n(x)+s x)
≤ α q(x, T (j−1)n(x)+s x), q(x, T n(x) x), q(x, T n(x) T (j−1)n(x)+s x),
q(T n(x) x, T (j−1)n(x)+s x), q(T n(x) T (j−1)n(x)+s x, T (j−1)n(x)+s x)
= α uj−1 , q(x, T n(x) x), q(x, T j n(x)+s x), q(T n(x) x, T (j−1)n(x)+s x),
q(T j n(x)+s x, T (j−1)n(x)+s x)
≤ α uj−1 , h, uj , q(T n(x) x, T (j−1)n(x)+s x), q(T j n(x)+s x, T (j−1)n(x)+s x) .
(6)
Notice that
q(T n(x) x, T (j−1)n(x)+s x) ≤ q(T n(x) x, x) + q(x, T (j−1)n(x)+s x)
≤ M q(x, T n(x) x) + q(x, T (j−1)n(x)+s x)
≤ M h + uj−1 ≤ M c + c ≤ (M + 1) uj ,
and
q(T j n(x)+s x, T (j−1)n(x)+s x) ≤ q(T j n(x)+s x, x) + q(x, T (j−1)n(x)+s x)
≤ M q(x, T j n(x)+s x) + q(x, T (j−1)n(x)+s x) = M uj + uj−1 ≤ (M + 1) uj .
As α is nondecreasing on each variable, it follows from (6) that
q(T n(x) x, T n(x) T (j−1)n(x)+s x)
≤ α uj−1 , h, uj , q(T n(x) x, T (j−1)n(x)+s x), q(T j n(x)+s x, T (j−1)n(x)+s x)
≤ α (uj , uj , uj , (M + 1)uj , (M + 1)uj ) ≤ γ (uj ) .
Then (5) implies that
uj ≤ h + q(T n(x) x, T n(x) T (j−1)n(x)+s x) ≤ h + γ (uj )
Therefore, uj − γ (uj ) ≤ h, which contradicts (4) because we suppose that
uj ≥ c. Therefore, it is impossible to find such j, which proves that
q(x, T k n(x)+s x) = uk < c for all k ∈ {0, 1, 2, . . .} .
(7)
This proves that the set {q(x, T k n(x)+s x) : k ∈ N} is bounded, and varying s
on {0, 1, 2, . . . , n(x) − 1}, we conclude that Step 1 holds. Let define
Sx1 = sup{q(x, T k x) : k ∈ {0, 1, 2, . . .}} .
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Step 2. We claim that, for all x ∈ X, the set {q(T k x, x) : k ∈ N} is
bounded.
This step follows from the fact that q(T k x, x) ≤ M q(x, T k x) ≤ M · Sx1 for
all x ∈ X and all k ∈ N. Let define, for all x ∈ X,
Sx2 = sup{q(T k x, x) : k ∈ {0, 1, 2, . . .}}
and Sx = Sx1 + Sx2 .
As a consequence of Steps 1 and 2, we have proved that, for all x ∈ X, the
orbit {T k x : k ∈ N} is a q-bounded subset of X because
0
0
q(T k x, T k x) ≤ q(T k x, x) + q(x, T k x) ≤ Sx1 + Sx2 = Sx
for all k, k 0 ∈ N.
If x is a fixed point of T , then the existence of such kind of points is proved.
On the contrary case, if T x 6= x, then
Sx1 ≥ q(x, T x) > 0
and Sx2 ≥ q(T x, x) > 0.
(8)
The following steps have sense when we begin the process using a point x0 ∈ X
which is not a fixed point of T .
Step 3. Starting from an arbitrary point x0 ∈ X, we claim that the
iterative sequence {xk }k≥0 given by
xk+1 = T nk xk
for all k ∈ N
(where nk = n(xk ))
(9)
verifies the following properties
xk+1 = T n0 +n1 +...+nk x0 ,
xk+j = T
nk +nk+1 +...+nk+j−1
(10)
xk
for all k, j ∈ N.
(11)
Indeed, notice that
x 1 = T n0 x 0 ,
x2 = T n1 x1 = T n1 T n0 x0 = T n0 +n1 x0 ,
and, by induction methodology, we deduce that (10) holds. This means that
{xk }k≥0 is a subsequence of the orbit {T k x0 }∞
k=0 . To prove (11), we observe
that
xk+j = T n0 +n1 +...+nk+j−1 x0 = T nk +nk+1 +...+nk+j−1 T n0 +n1 +...+nk−1 x0
= T nk +nk+1 +...+nk+j−1 xk .
Step 4. We claim that {xk }k≥0 is a right-Cauchy sequence on (X, q).
Matkowski theorems in the context of quasi-metric spaces and
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If T x0 = x0 , then x0 is a fixed point of T (and the existence of such kind
of points is guaranteed). Assume that T x0 6= x0 . Let ε > 0 be arbitrary. By
(P4 ), limk→∞ γ k (Sx10 ) = 0. Hence, there exists k0 ∈ N such that
γ k (Sx10 ) < ε for all k ≥ k0 .
(12)
Let k, j ∈ N be arbitrary integers such that k ≥ k0 and j ≥ 1. Denote by s0
the integer s0 = nk + nk+1 + . . . + nk+j−1 and let s1 ∈ {s0 , nk−1 , s0 + nk−1 }
be the appropriate index such that
max q(xk−1 , T s0 xk−1 ), q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T s0 +nk−1 xk−1 )
= q(xk−1 , T s1 xk−1 ).
(13)
By using (10) and (11), we get that
q(xk , xk+j ) = q(T nk−1 xk−1 , T nk−1 +nk +nk+1 +...+nk+j−1 xk−1 )
= q(T nk−1 xk−1 , T nk−1 T nk +nk+1 +...+nk+j−1 xk−1 )
= q(T nk−1 xk−1 , T nk−1 T s0 xk−1 ).
Using the contractivity condition (3) with y = T s0 xk−1 , we observe that
q(xk , xk+j ) = q(T nk−1 xk−1 , T nk−1 T s0 xk−1 )
≤ α q (xk−1 , T s0 xk−1 ) , q(xk−1 , T n(xk−1 ) xk−1 ), q(xk−1 , T n(xk−1 ) T s0 xk−1 ),
q(T n(xk−1 ) xk−1 , T s0 xk−1 ), q(T n(xk−1 ) T s0 xk−1 , T s0 xk−1 )
≤ α q (xk−1 , T s0 xk−1 ) , q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T s0 +nk−1 xk−1 ),
q(T nk−1 xk−1 , T s0 xk−1 ), q(T nk−1 +s0 xk−1 , T s0 xk−1 ) .
(14)
Taking into account (13), we observe that
q(T nk−1 xk−1 , T s0 xk−1 ) ≤ q(T nk−1 xk−1 , xk−1 ) + q(xk−1 , T s0 xk−1 )
≤ M q(xk−1 , T nk−1 xk−1 ) + q(xk−1 , T s0 xk−1 ) ≤ (M + 1) q(xk−1 , T s1 xk−1 )
and
q(T nk−1 +s0 xk−1 , T s0 xk−1 ) ≤ q(T nk−1 +s0 xk−1 , xk−1 ) + q(xk−1 , T s0 xk−1 )
≤ M q(xk−1 , T nk−1 +s0 xk−1 ) + q(xk−1 , T s0 xk−1 )
≤ (M + 1) q(xk−1 , T s1 xk−1 ).
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As α is nondecreasing on each argument, it follows from (14) that
q(xk , xk+j ) ≤ α q (xk−1 , T s0 xk−1 ) , q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T s0 +nk−1 xk−1 ),
q(T nk−1 xk−1 , T s0 xk−1 ), q(T nk−1 +s0 xk−1 , T s0 xk−1 )
≤ α (q(xk−1 , T s1 xk−1 ), q(xk−1 , T s1 xk−1 ), q(xk−1 , T s1 xk−1 ),
(M + 1) q(xk−1 , T s1 xk−1 ), (M + 1) q(xk−1 , T s1 xk−1 ))
≤ γ(q(xk−1 , T s1 xk−1 )).
Using s2 ∈ {s1 , nk−2 , s1 + nk−2 } be the appropriate index such that
max q(xk−2 , T s1 xk−2 ), q(xk−2 , T nk−2 xk−2 ), q(xk−2 , T s1 +nk−2 xk−2 )
= q(xk−2 , T s2 xk−2 ),
and repeating the previous argument, we deduce that
q(xk−1 , T s1 xk−1 ) ≤ γ(q(xk−2 , T s2 xk−2 )).
Hence, as γ is nondecreasing, we have that
q(xk , xk+j ) ≤ γ(q(xk−1 , T s1 xk−1 )) ≤ γ 2 (q(xk−2 , T s2 xk−2 ))
≤ . . . ≤ γ k (q(x0 , T sk x0 ))
for appropriate indices s0 , s1 , . . . , sk ∈ N. Since q(x0 , T sk x0 ) ≤ Sx10 and γ k is
nondecreasing, if we take k ≥ k0 , it follows from (12) that
q(xk , xk+j ) ≤ γ k (q(x0 , T sk x0 )) ≤ γ k (Sx10 ) < ε.
As k ≥ k0 and j ≥ 1 are arbitrary, we conclude that {xk }k≥0 is a right-Cauchy
sequence on (X, q).
Step 5. We claim that {xk }k≥0 is a left-Cauchy sequence on (X, q).
This step follows from Step 4 and item 3 of Lemma 3.3.
Step 6. {xk }k≥0 converges to some a ∈ X.
Since {xk }k≥0 is both left- and right-Cauchy, Remark 2.1 guarantees that
it is a Cauchy sequence on (X, q). As it is complete, there exists a ∈ X such
that {xk } → a, which means that
lim q(a, xk ) = lim q(xk , a) = 0.
k→∞
k→∞
Step 7. We claim that
lim q(xk , T n(a) xk ) = 0.
k→∞
(15)
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Indeed, using the contractivity condition (3), we have that, for all k ≥ 1,
q(xk , T n(a) xk ) = q(T nk−1 xk−1 , T n(a) T nk−1 xk−1 ) = q(T nk−1 xk−1 , T nk−1 T n(a) xk−1 )
≤ α q(xk−1 , T n(a) xk−1 ), q(xk−1 , T n(xk−1 ) xk−1 ), q(xk−1 , T n(xk−1 ) T n(a) xk−1 ),
q(T n(xk−1 ) xk−1 , T n(a) xk−1 ), q(T n(xk−1 ) T n(a) xk−1 , T n(a) xk−1 )
= α q(xk−1 , T n(a) xk−1 ), q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T nk−1 +n(a) xk−1 ),
(16)
q(T nk−1 xk−1 , T n(a) xk−1 ), q(T nk−1 +n(a) xk−1 , T n(a) xk−1 ) .
Let r1 ∈ {n(a), nk−1 , nk−1 + n(a)} be an appropriate index such that
n
o
max q(xk−1 , T n(a) xk−1 ), q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T n(a)+nk−1 xk−1 )
= q(xk−1 , T r1 xk−1 ).
Since
q(T nk−1 xk−1 , T n(a) xk−1 ) ≤ q(T nk−1 xk−1 , xk−1 ) + q(xk−1 , T n(a) xk−1 )
≤ M q(xk−1 , T nk−1 xk−1 ) + q(xk−1 , T n(a) xk−1 ) ≤ (M + 1) q(xk−1 , T r1 xk−1 ),
and
q(T nk−1 +n(a) xk−1 , T n(a) xk−1 )
≤ q(T nk−1 +n(a) xk−1 , xk−1 ) + q(xk−1 , T n(a) xk−1 )
≤ M q(xk−1 , T nk−1 +n(a) xk−1 ) + q(xk−1 , T n(a) xk−1 )
≤ (M + 1) q(xk−1 , T r1 xk−1 ),
then it follows from (16) that
q(xk , T n(a) xk ) ≤ α q(xk−1 , T n(a) xk−1 ), q(xk−1 , T nk−1 xk−1 ), q(xk−1 , T nk−1 +n(a) xk−1 ),
q(T nk−1 xk−1 , T n(a) xk−1 ), q(T nk−1 +n(a) xk−1 , T n(a) xk−1 )
≤ α (q (xk−1 , T r1 xk−1 ) , q(xk−1 , T r1 xk−1 ), q(xk−1 , T r1 xk−1 ),
(M + 1) q(xk−1 , T r1 xk−1 ), (M + 1) q(xk−1 , T r1 xk−1 ))
≤ γ(q (xk−1 , T r1 xk−1 )).
Repeating this argument and taking into account that γ is nondecreasing, it
yields that, for all k ∈ N,
q(xk , T n(a) xk ) ≤ γ(q (xk−1 , T r1 xk−1 )) ≤ γ 2 (q (xk−2 , T r2 xk−2 )) ≤ . . .
≤ γ k (q (x0 , T rk x0 )) ≤ γ k (Sx10 ).
By using (P4 ), we conclude that Step 7 holds.
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Step 8. We claim that
lim q(T n(a) xk , xk ) = 0.
k→∞
(17)
This step follows from Step 7 and from item 4 of Lemma 3.3.
Step 9. We claim that T n(a) a = a, that is, a is a fixed point of T n(a) .
We reason by contradiction. Assume that T n(a) a 6= a and let
ε1 = q(a, T n(a) a) > 0
and ε2 = q(T n(a) a, a) > 0.
Without loss of generality, suppose that ε1 ≤ ε2 (the contrary case is similar).
By item 1 of Lemma 3.2,
γ(ε1 ) < ε1 .
Let k0 ∈ N be such that
n
o
max q (a, xk ) , q (xk , a) , q(T n(a) xk , xk ), q(xk , T n(a) xk )
≤
ε1 − γ(ε1 )
< ε1
4
for all k ≥ k0 .
By using the triangle inequality,
ε2 = q(T n(a) a, a) ≤ q(T n(a) a, T n(a) xk ) + q(T n(a) xk , xk ) + q (xk , a)
≤ q(T n(a) a, T n(a) xk ) +
ε1 − γ(ε1 )
.
2
(18)
Since
q(T n(a) a, xk ) ≤ q(T n(a) a, a)+q(a, xk ) ≤ M q(a, T n(a) a)+q(a, xk ) ≤ (M +1)ε1
and
q(a, T n(a) xk ) ≤ q(a, xk ) + q(xk , T n(a) xk )
≤
ε1 − γ(ε1 ) ε1 − γ(ε1 )
ε1 − γ(ε1 )
+
=
< ε1 ,
4
4
2
the contractivity condition (3) guarantees that
q(T n(a) a, T n(a) xk )
≤ α q (a, xk ) , q(a, T n(a) a), q(a, T n(a) xk ), q(T n(a) a, xk ), q(T n(a) xk , xk )
≤ α (ε1 , ε1 , ε1 , (M + 1)ε1 , ε1 ) ≤ α (ε1 , ε1 , ε1 , (M + 1)ε1 , (M + 1)ε1 )
≤ γ(ε1 ).
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From (18), it yields
ε1 − γ(ε1 )
2
ε1 − γ(ε1 )
ε1 + γ(ε1 )
ε1 + ε1
=
<
= ε1 ≤ ε2 ,
≤ γ(ε1 ) +
2
2
2
ε2 ≤ q(T n(a) a, T n(a) xk ) +
which is a contradiction. As a consequence, T n(a) a = a.
Step 10. We claim that a is the unique fixed point of T n(a) .
Suppose, on the contrary, that there exists b ∈ X, with b 6= a, such that
T n(a) b = b. Taking (3) into account, we get that
q(a, b) = q(T n(a) a, T n(a) b)
≤ α q (a, b) , q(a, T n(a) a), q(a, T n(a) b), q(T n(a) a, b), q(T n(a) b, b)
= α (q (a, b) , q(a, a), q(a, b), q(a, b), q(b, b)) = α (q (a, b) , 0, q(a, b), q(a, b), 0)
≤ α (q (a, b) , q (a, b) , q(a, b), (M + 1) q(a, b), (M + 1) q(a, b))
≤ γ(q(a, b)) < q(a, b),
which is a contradiction. Hence, a is the unique fixed point of T n(a) .
Step 11. We claim that a is a fixed point of T .
By step 9, T n(a) (T a) = T (T n(a) a) = T a, which means that T a is also a
fixed point of T n(a) . And by Step 10, T a = a.
Step 12. We claim that, for each x ∈ X, we have that lim q(a, T k x) = 0.
k→∞
For this purpose, we fix x ∈ X and let s ∈ {0, 1, 2, . . . , n(a) − 1} be arbitrary. Let define
bk = q(a, T k n(a)+s x)
for all k = {0, 1, 2, . . .} .
Let show that bk ≤ bk−1 for all k ≥ 1 reasoning by contradiction. Assume
that there exists some k0 ∈ N such that bk0 > bk0 −1 . Therefore,
bk0 = q(a, T k0 n(a)+s x) = q(T n(a) a, T n(a) T (k0 −1)n(a)+s x)
≤ α q(a, T (k0 −1)n(a)+s x), q(a, T n(a) a), q(a, T n(a) T (k0 −1)n(a)+s x),
q(T n(a) a, T (k0 −1)n(a)+s x), q(T n(a) T (k0 −1)n(a)+s x, T (k0 −1)n(a)+s x)
≤ α bk0 −1 , 0, q(a, T k0 n(a)+s x),
q(a, T (k0 −1)n(a)+s x), q(T k0 n(a)+s x, T (k0 −1)n(a)+s x)
= α bk0 −1 , 0, bk0 , bk0 −1 , q(T k0 n(a)+s x, T (k0 −1)n(a)+s x) .
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Since
q(T k0 n(a)+s x, T (k0 −1)n(a)+s x) ≤ q(T k0 n(a)+s x, a) + q(a, T (k0 −1)n(a)+s x)
≤ M q(a, T k0 n(a)+s x) + q(a, T (k0 −1)n(a)+s x) = M bk0 + bk0 −1
≤ (M + 1) bk0 ,
taking into account that bk0 > bk0 −1 ≥ 0, it follows that
bk0 ≤ α bk0 −1 , 0, bk0 , bk0 −1 , q(T k0 n(a)+s x, T (k0 −1)n(a)+s x)
≤ α (bk0 , bk0 , bk0 , (M + 1) bk0 , (M + 1) bk0 ) ≤ γ(bk0 ) < bk0 ,
which is a contradiction. This proves that
bk ≤ bk−1
for all k ≥ 1.
Repeating the previous argument, for all k ∈ N,
bk = q(a, T k n(a)+s x) = q(T n(a) a, T n(a) T (k−1)n(a)+s x)
≤ α q(a, T (k−1)n(a)+s x), q(a, T n(a) a), q(a, T n(a) T (k−1)n(a)+s x),
q(T n(a) a, T (k−1)n(a)+s x), q(T n(a) T (k−1)n(a)+s x, T (k−1)n(a)+s x)
≤ α bk−1 , 0, q(a, T k n(a)+s x),
q(a, T (k−1)n(a)+s x), q(T k n(a)+s x, T (k−1)n(a)+s x)
= α bk−1 , 0, bk , bk−1 , q(T k n(a)+s x, T (k−1)n(a)+s x) ,
where
q(T k n(a)+s x, T (k−1)n(a)+s x) ≤ q(T k n(a)+s x, a) + q(a, T (k−1)n(a)+s x)
≤ M q(a, T k n(a)+s x) + q(a, T (k−1)n(a)+s x) = M bk + bk−1 ≤ (M + 1) bk−1 .
Thus,
bk ≤ α bk−1 , 0, bk , bk−1 , q(T k n(a)+s x, T (k−1)n(a)+s x)
≤ α (bk−1 , bk−1 , bk , (M + 1) bk−1 , (M + 1) bk−1 ) ≤ γ(bk−1 ).
As γ is nondecreasing,
bk ≤ γ(bk−1 ) ≤ γ 2 (bk−2 ) ≤ . . . ≤ γ k (b0 ) = γ k (q(a, T s x)).
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Since a, x and s are fixed, then property (P4 ) implies that {bk } → 0, that is
{q(a, T k n(a)+s x)} → 0.
Varying s ∈ {0, 1, 2, . . . , n(a) − 1}, we conclude that Step 12 holds.
Step 13. We claim that, for each x ∈ X, we have that lim q(T k x, a) = 0.
k→∞
This step follows from Step 12 and from item 4 of Lemma 3.3. Joining
Steps 12 and 13, we have proved that {T k x} → a for all x ∈ X.
Step 14. We shall show that T n(a) is continuous at a.
To prove this assertion, we take an arbitrary sequence {yk } ⊆ X that
converges to a, that is,
lim q (yk , a) = lim q (a, yk ) = 0.
k→∞
k→∞
(19)
Let show that
q(a, T n(a) yk ) ≤ q(a, yk )
for all k ∈ N
(20)
by contradiction.
Assume that there exists some k ∈ N such that
q(a, T n(a) yk ) > q(a, yk ). Again by (3), we have that
q(a, T n(a) yk ) = q(T n(a) a, T n(a) yk )
≤ α q (a, yk ) , q(a, T n(a) a), q(a, T n(a) yk ), q(T n(a) a, yk ), q(T n(a) yk , yk )
≤ α q (a, yk ) , 0, q(a, T n(a) yk ), q(a, yk ), q(T n(a) yk , yk ) .
Since
q(T n(a) yk , yk ) ≤ q(T n(a) yk , a) + q(a, yk )
≤ M q(a, T n(a) yk ) + q(a, yk ) ≤ (M + 1)q(a, T n(a) yk ),
therefore
q(a, T n(a) yk ) ≤ α q (a, yk ) , 0, q(a, T n(a) yk ), q(a, yk ), q(T n(a) yk , yk )
≤ α q(a, T n(a) yk ), q(a, T n(a) yk ), q(a, T n(a) yk ),
(M + 1)q(a, T n(a) yk ), (M + 1)q(a, T n(a) yk )
≤ γ(q(a, T n(a) yk )) < q(a, T n(a) yk ),
which is a contradiction. As a consequence, (20) holds, and by (19),
{q(a, T n(a) yk )} → 0. Moreover, as q(T n(a) yk , a) ≤ M q(a, T n(a) yk ) for all
k ∈ N, then also {q(T n(a) yk , a)} → 0. Therefore, {T n(a) yk } → a = T n(a) a,
which means that T n(a) is continuous at x = a. This finishes the proof.
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Corollary 3.1. Let (X, q) be a complete M -symmetric quasi-metric space, let
T : X → X be a mapping and let ϕ ∈ FMat be a Matkowski function. Assume
that, for each x ∈ X, there is a positive integer n = n(x) such that
for all y ∈ X,
q(T n(x) x, T n(x) y) ≤ ϕ(q(x, y)).
(21)
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
5
Proof. Given ϕ ∈ FMat , let define αϕ : [0, ∞) → [0, ∞) for all t1 , t2 , t3 , t4 , t5 ∈
[0, ∞) by
αϕ (t1 , t2 , t3 , t4 , t5 ) = ϕ(t1 ).
Then αϕ is nondecreasing on each argument and
αϕ (t, t, t, (M + 1) t, (M + 1) t) = ϕ(t).
Then (αϕ , ϕ, M ) is a Matkowski’s triple and
q(T n(x) x, T n(x) y) ≤ ϕ(q(x, y))
= αϕ q (x, y) , q(x, T n(x) x), q(x, T n(x) y), q(T n(x) x, y), q(T n(x) y, y) .
This means that Theorem 3.1 is applicable.
In the following result, the integer n = n(x) is constant.
Corollary 3.2. Let (X, q) be a complete M -symmetric quasi-metric space and
let T : X → X be a mapping. Suppose that there exists a Matkowski’s triple
(α, γ, M ) and a positive integer number n such that, for all x, y ∈ X,
q(T n x, T n y) ≤ α (q (x, y) , q(x, T n x), q(x, T n y), q(T n x, y), q(T n y, y))
Then T has a unique fixed point a ∈ X. Furthermore, for each x ∈ X,
lim T k x = a and T n is continuous at a.
k→∞
If we take λ ∈ [0, 1) and ϕλ (t) = λ t for all t ≥ 0, then ϕλ ∈ FMat and we
get the following result.
Corollary 3.3. Let (X, q) be a complete M -symmetric quasi-metric space, let
T : X → X be a mapping and let λ ∈ [0, 1). Assume that, for each x ∈ X,
there is a positive integer n = n(x) such that
for all y ∈ X,
q(T n(x) x, T n(x) y) ≤ λ q(x, y).
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
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If we take ϕ(t) = t/ (1 + t) for all t ≥ 0, then ϕ ∈ FMat and we get the
following result.
Corollary 3.4. Let (X, q) be a complete M -symmetric quasi-metric space and
let T : X → X be a mapping. Assume that, for each x ∈ X, there is a positive
integer n = n(x) such that
for all y ∈ X,
q(T n(x) x, T n(x) y) ≤
q(x, y)
.
1 + q(x, y)
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
3.3
The Matkowski’s theorem using a symmetric contractivity condition on quasi-metric spaces
In this subsection, we show how it is possible to consider a version of Theorem
1.1 on quasi metric spaces using a symmetric contractivity condition, like in
the following result.
Theorem 3.2. Let (X, q) be a complete quasi-metric space, let T : X → X be
a mapping and let ϕ ∈ FMat be a Matkowski function. Assume that, for each
x ∈ X, there is a positive integer n = n(x) such that
for all y ∈ X,
q(T n(x) x, T n(x) y) ≤ ϕ(q(x, y))
q(T n(x) y, T n(x) x) ≤ ϕ(q(y, x)).
and
(22)
(23)
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
Although Theorem 3.2 seems to be an extension of Theorem 1.1, actually,
it is not a true generalization. In fact, we are going to show that Theorem 1.1
and Theorem 3.2 are, indeed, equivalent.
Lemma 3.4. Theorem 1.1 follows from Theorem 3.2.
Proof. If q is a metric on X, then both conditions (22) and (23) are equivalent
to (2). Then, if we assume that Theorem 3.2 holds, it is evident that Theorem
1.1 also holds.
Lemma 3.5. Theorem 3.2 follows from Theorem 1.1.
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Proof. Assume that Theorem 1.1 holds, and we are going to show that Theorem 3.2 also holds. Let (X, q) be a complete quasi-metric space, let T : X → X
be a mapping and let ϕ ∈ FMat be a Matkowski function. Assume that, for
each x ∈ X, there is a positive integer n = n(x) such that (22)-(23) holds. Let
define dq : X × X → [0, ∞) by
dq (x, y) = max {q (x, y) , q (y, x)}
for all x, y ∈ X.
As (X, q) is a complete quasi-metric space, it is well-know that dq is a complete
metric on X. Furthermore, given x ∈ X, let n(x) ∈ N be such that (22)-(23)
holds. Then, taking into account that ϕ is a nondecreasing function, we have,
for all y ∈ X,
n
o
dq (T n(x) x, T n(x) y) = max q(T n(x) x, T n(x) y), q(T n(x) y, T n(x) x)
≤ max {ϕ(q(x, y)), ϕ(q(y, x))}
= ϕ (max {q(x, y), q(y, x)})
= ϕ(dq (x, y)).
Hence, T verifies the contractivity condition of Theorem 1.1. Such theorem
guarantees that T has a unique fixed point a ∈ X and that lim T k x = a for
k→∞
each x ∈ X.
In our main theorems, we have used that ϕ is a nondecreasing function
satisfying (ϕ3 ) = (P4 ) and (ϕ6 ) = (P3 ). By Corollary 2.1, we can replace
these conditions by another ones.
Corollary 3.5. Corollary 3.1 and Theorem 3.2 also hold if ϕ is a nondecreasing function satisfying (ϕ4 ) and (ϕ5 ).
If we take λ ∈ [0, 1) and ϕλ (t) = λ t for all t ≥ 0, then ϕλ ∈ FMat and we
get the following result.
Corollary 3.6. Let (X, q) be a complete quasi-metric space, let T : X → X be
a mapping and let λ ∈ [0, 1). Assume that, for each x ∈ X, there is a positive
integer n = n(x) such that
for all y ∈ X,
q(T n(x) x, T n(x) y) ≤ λ q(x, y)
q(T
n(x)
y, T
n(x)
and
x) ≤ λ q(y, x).
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
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If we take ϕ(t) = t/ (1 + t) for all t ≥ 0, then ϕ ∈ FMat and we get the
following result.
Corollary 3.7. Let (X, q) be a complete quasi-metric space and let T : X →
X be a mapping. Assume that, for each x ∈ X, there is a positive integer
n = n(x) such that
for all y ∈ X,
q(x, y)
1 + q(x, y)
q(y, x)
q(T n(x) y, T n(x) x) ≤
.
1 + q(y, x)
q(T n(x) x, T n(x) y) ≤
and
Then T has a unique fixed point a ∈ X. Moreover, for each x ∈ X, we have
that lim T k x = a and T n(a) is continuous at a.
k→∞
4
Consequences: Fixed Point Results on G-Metric Spaces
In this section, we particularize the previous results to the setting of G-metric
spaces, and we show that some existing fixed point results in the context of Gmetric spaces can be easily deduced from our main theorems. For the sake of
completeness, we collect here some definitions and basic result about G-metric
spaces (for more details, see e.g. [3]-[18]).
In 2003, Mustafa and Sims [12] proved that most of the claims concerning
the topological properties of D-metrics were incorrect. In order to repair these
drawbacks, they gave a more appropriate notion of generalized metrics, called
G-metrics.
Definition 4.1. (Mustafa and Sims [12]) A G-metric space is a pair (X, G)
where X is a nonempty set and G : X × X × X → [0, ∞) is a function such
that, for all x, y, z, a ∈ X, the following conditions are fulfilled:
(G1 )
G(x, y, z) = 0
if x = y = z;
(G2 )
G(x, x, y) > 0
for all x, y ∈ X with x 6= y;
(G3 )
G(x, x, y) ≤ G(x, y, z)
for all x, y, z ∈ X with z 6= y;
(G4 )
G(x, y, z) = G(x, z, y) = G(y, z, x) = . . . ( symmetry in all three variables);
(G5 )
G(x, y, z) ≤ G(x, a, a) + G(a, y, z)
( rectangle inequality).
In such a case, the function G is called a G-metric on X.
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330
Example 4.1. If X is a non-empty subset of R, then the function G : X ×
X × X → [0, ∞), given by
G (x, y, z) = |x − y| + |x − z| + |y − z|
for all x, y, z ∈ X,
is a G-metric on X.
One of the most useful properties of G-metrics is the following one.
Lemma 4.1. If (X, G) is a G-metric space, then
G(x, y, y) ≤ 2 G(y, x, x)
for all x, y ∈ X.
Definition 4.2. Let (X, G) be a G-metric space, let x ∈ X be a point and let
{xm } ⊆ X be a sequence. We say that:
G
• {xm } G-converges to x, and we write {xn } −→ x or {xn } → x, if
limm,m0 →∞ G(xm , xm0 , x) = 0, that is, for all ε > 0 there exists m0 ∈ N
verifying that G(xm , xm0 , x) < ε for all m, m0 ∈ N such that m, m0 ≥ m0
(in such a case, x if the G-limit of {xm });
• {xm } is G-Cauchy if limm,m0 ,m00 →∞ G(xm , xm0 , xm00 ) = 0, that is, for
all ε > 0 there exists m0 ∈ N verifying that G(xm , xm0 , xm00 ) < ε for all
m, m0 , m00 ∈ N such that m, m0 , m00 ≥ m0 .
• (X, G) is complete if every G-Cauchy sequence in X is G-convergent in
X.
Theorem 4.1. Let (X, G) be a G-metric space and let qG : X × X → [0, ∞)
be the function defined by qG (x, y) = G(x, y, y) for all x, y ∈ X. Then:
1. (X, qG ) is a 2-symmetric quasi-metric space;
2. {xn } ⊆ X is G-convergent to x ∈ X if, and only if, {xn } is convergent
to x in (X, qG );
3. {xn } ⊆ X is G-Cauchy if, and only if, {xn } is Cauchy in (X, qG );
4. (X, G) is G-complete if, and only if, (X, qG ) is complete.
0
Lemma 4.2. The previous theorem also holds if we replace qG by qG
, defined
0
by qG (x, y) = G(x, x, y) for all x, y ∈ X.
The following one is the particularization of Theorem 3.1 to quasi-metric
spaces of the form (X, qG ) given in Theorem 4.1.
Matkowski theorems in the context of quasi-metric spaces and
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331
Corollary 4.1. Let (X, G) be a complete G-metric space and let T : X → X
be a mapping. Suppose that there exists a Matkowski’s triple (α, γ, 2) verifying
the following property:
• for every x ∈ X, there is a positive integer n = n(x) such that, for all
y ∈ X,
G(T n(x) x, T n(x) y, T n(x) y) ≤ α G (x, y, y) , G(x, T n(x) x, T n(x) x),
G(x, T n(x) y, T n(x) y), G(T n(x) x, y, y), G(T n(x) y, y, y) .
Then T has a unique fixed point a ∈ X. Furthermore, for each x ∈ X,
lim T k x = a and T n(a) is continuous at a.
k→∞
0
as in Lemma 4.2 is given in the following stateThe same result using qG
ment.
Corollary 4.2. Let (X, G) be a complete G-metric space and let T : X → X
be a mapping. Suppose that there exists a Matkowski’s triple (α, γ, 2) verifying
the following property:
• for every x ∈ X, there is a positive integer n = n(x) such that, for all
y ∈ X,
G(T n(x) x, T n(x) x, T n(x) y) ≤ α G (x, x, y) , G(x, x, T n(x) x),
G(x, x, T n(x) y), G(T n(x) x, T n(x) x, y), G(T n(x) y, T n(x) y, y) .
Then T has a unique fixed point a ∈ X. Furthermore, for each x ∈ X,
lim T k x = a and T n(a) is continuous at a.
k→∞
As a consequence, the main result in [2] can be seen as a simple consequence
of the previous results.
Theorem 4.2. Theorem 1.2 immediately follows from Corollary 3.1.
Proof. It is sufficient to take qG (x, y) = G(x, x, y) for all x, y ∈ X in Corollary
3.1 together with Theorem 4.1. Notice that the conditions (ϕ3 )&(ϕ6 ) and
(ϕ4 )&(ϕ5 ) are equivalent due to Corollary 2.1.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Matkowski theorems in the context of quasi-metric spaces and
consequences on G-metric spaces
332
Authors’ contributions
All authors contributed equally and significantly in writing this article. All
authors read and approved the final manuscript.
Acknowledgements
The second author has been partially supported by Junta de Andalucı́a by
project FQM-268 of the Andalusian CICYE. The third author would like to
extend his sincere appreciation to the Deanship of Scientific Research at King
Saud University for its funding of this research through the Research Group
Project no RGP-VPP-237.
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Erdal Karapınar,
Department of Mathematics,
Atilim University,
06836, İncek, Ankara, Turkey.
Email: erdal.karapinar@atilim.edu.tr
Antonio-Francisco Roldán-López-de-Hierro,
Department of Quantitative Methods for Economics and Business,
Faculty of Economic and Business Sciences,
University of Granada,
Campus Universitario de La Cartuja, 18011, Granada, Spain.
Email: aroldan@ugr.es, afroldan@ujaen.es
Bessem Samet,
Department of Mathematics,
King Saud University, Saudi Arabia.
Email: bsamet@ksu.edu.sa
Matkowski theorems in the context of quasi-metric spaces and
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